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Life Cycle Reliability and Safety
Engineering
ISSN 2520-1352
Life Cycle Reliab Saf Eng
DOI 10.1007/s41872-020-00159-4
Reliability modelling and analysis of client–
server system using Gumbel–Hougaard
family copula
Ibrahim Yusuf, Abdulkareem Lado
Ismail, Muhammad Auwal Lawan,
U.A.Ali & Sufi Nasir
1 23
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Vol.:(0123456789)
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Life Cycle Reliability and Safety Engineering
https://doi.org/10.1007/s41872-020-00159-4
ORIGINAL RESEARCH
Reliability modelling andanalysis ofclient–server system using
Gumbel–Hougaard family copula
IbrahimYusuf1 · AbdulkareemLadoIsmail2· MuhammadAuwalLawan3· U.A.Ali4· SuNasir5
Received: 8 May 2020 / Accepted: 25 November 2020
© Society for Reliability and Safety (SRESA) 2021
Abstract
The present paper deals with the reliability analysis of a computer network system as series parallel system consisting of
two subsystems. Subsystem I consist of four homogeneous clients while subsystem II consist of two homogeneous servers.
Both clients and servers have exponential failure whereas repairs follow two types of distributions that are general and Gum-
bel–Hougaard family copula. The network under consideration has three states: normal, partial failure state and complete
failure state. From the transition diagram, the system of first-order partial differential equations is derived and solved using
a supplementary variable technique and Laplace transforms. The system is analyzed using Laplace transforms to solve the
mathematical equations. The results obtained are presented in tables and graphs. Some important measures of reliability such
as availability of system, reliability of the system, MTTF, sensitivity and cost analysis have been discussed. Some particular
cases have also been derived and examined to see the practical effect of the model.
Keywords Reliability· Performance· Availability· Gumbel–Hougaard family copula· Mean time to failure· Distributed
database servers (DDS I and DDS II)
1 Introduction
The computer network consists of devices joined together to
perform a number of applications or functions or to evalu-
ate different tasks. Computer network consists of devices
such as computers, load balancing devices, output devices,
and servers that communicate with clients to share informa-
tion through request and response. The application of the
computer network finds its way into human endeavors such
as education, banking, industrial, and manufacturing set-
tings, etc. Computer network systems must, therefore, run
a failure free of charge to improve reliability, availability,
lower maintenance costs, quality of output, revenue, quality
of service and safety in order to meet the daily needs of the
human race. With the advancement in the field of computer
technology, the use of the computer network is becoming a
subject for discussion. Computer network is a client server
application that has been used as a medium to share or dis-
tribute information locally or internationally. The client is
a computer hardware that sends a request to the server that
is connected to it, while the server is also a computer that
accepts a request from the client, processes it and responds
to a request from the client. In a computer network with mul-
tiple clients and multiple servers, clients can send request
* Ibrahim Yusuf
iyusuf.mth@buk.edu.ng
Abdulkareem Lado Ismail
ladgetso@gmail.com
Muhammad Auwal Lawan
mukhazah3@gmail.com
U. A. Ali
ubahamad@yahoo.co.uk
Sufi Nasir
sufinasir@rocketmail.com
1 Department ofMathematical Sciences, Bayero University,
Kano, Nigeria
2 Department ofMathematics, Kano State College
ofEducation andPreliminary Studies, Kano, Nigeria
3 Department ofMathematics, Kano University ofScience
andTechnology, Wudil, Kano, Nigeria
4 Department ofMathematics, Federal University, Dutse,
Nigeria
5 Department ofComputer Science, Bayero University, Kano,
Nigeria
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Life Cycle Reliability and Safety Engineering
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services simultaneously and instantly to servers on the same
network as a local area network or a wide area network.
With rapid progress in the field of computer network
technology, performance analysis and evaluation, quality
assurance, network availability and reliability have become a
critical issue. Various authors have studied some remarkable
works in the field of system performance analysis and evalu-
ation of computer systems, industrial and manufacturing sys-
tems, etc. Alzubaidi etal. (2014) presented an enhancement
of the ICMP protocol to reduce IP over Ethernet architec-
ture. Gahlot etal. (2018) presented a performance assess-
ment of the serial system under different failure and repair
disciplines using copula linguistics. Garg (2013) analyzed
the performance of the repairable complex systems using
a PSO and a fuzzy confidence interval method. Ibrahim
etal. (2017) presented the performance of a complex sys-
tem containing two copulated subsystems in series. Kumar
etal. (2015) analyzed the performance of a computer system
with imperfect fault detection of hardware. Lado and Singh
(2019) presented the performance assessment of complex
repairable system consisting of two subsystems in series
using copula. Malik and Tewari (2018) presented perfor-
mance modelling and maintenance priorities decision for
water system of a coal based thermal power plant. Singh
and Singh (2015) analyzed the performance of three unit
redundant system with switch and human failure. Singh
etal. (2016) presented the performance of complex system
in series configuration under different failure and repair dis-
cipline using copula. Singh etal. (2020) presented the per-
formance of a complex repairable system with two subsys-
tems in series configuration with an imperfect switch. Gahlot
etal. (2018) analysed the performance of repairable system
in series configuration under different types of failure and
repair policies using Copula Linguistics. Gahlot etal. (2019)
analysed a two units’ complex repairable system with Switch
and Human failure using Copula Approach. Abubakar and
Singh (2019) addressed the performance of assessment of
African textile manufacturers, LTD, in Kano state, Nigeria,
through multi failure and repair using copula. Singh etal.
(2013) discussed availability, MTTF and cost analysis of a
system having two units in a series configuration with the
controller. Yusuf etal. (2019) performed reliability assess-
ment of a repairable system under online and offline preven-
tive maintenance. Yusuf etal. (2020) dealt with performance
analysis of multi computer system with three subsystems
in series. Isa etal. (2020) presented cost benefit analysis of
three different series–parallel dynamo configurations. Lado
etal. (2018) presented and cost assessment of repairable
complex system with two subsystems connected in series
configuration. Singh and Ayagi (2018) focused on stochas-
tic analysis of a complex system under preemptive resume
repair policy using Gumbel-Hougaard family of copula.
Due to their importance in communications, industrial
and manufacturing settings, banking, education, medical
systems, etc., many researchers have worked extensively
in the field of reliability and availability enhancement of
computer science and the computer network by obtaining
their reliability models to evaluate their performance under
different types of failure and repair policies. For instance
Anand and Malik (2012) analyzed a computer system with
arbitrary distributions for hardware and software replace-
ment time and priority to repair of hardware over software
replacement. Kumar and Malik (2012) presented economic
analysis of a computer system with software up gradation
and priority to hardware repair replacement. Jian and Shaop-
ing (2007) presented an integrated availability model based
on performance of computer networks. Kumar and Malik
(2012) presented stochastic modeling of a computer sys-
tem with priority to preventive maintenance over software
replacement subject to maximum operation and repair times.
Kumar etal. (2013) presented stochastic modeling of a com-
puter system with priority to Up-gradation of software over
hardware repair activities. Kumar and Malik (2014) dealt
with reliability modelling of a computer system with prior-
ity to hardware repair over replacement of hardware and
up-gradation of software subject to MOT and MRT. Kumar
etal. (2015) analyzed the performance of a computer system
with imperfect fault detection of hardware. Malik and Anand
(2010) presented the reliability and economic analysis of
a computer system with independent hardware and soft-
ware failures. Malik and Sureria (2012) presented analysis
of a computer system with hardware repair and software
replacement. Malik (2013) presented reliability modelling of
a computer system with preventive maintenance and priority.
Malik and Munday (2014) dealt with stochastic modelling of
a computer system with hardware redundancy. Munday and
Malik (2016) dealt with stochastic analysis of a computer
system with software redundancy.
Due to their importance in human endeavors, a wide
range of work in the field of reliability and accessibility of
computer systems and computer networks has been studied
by the above-mentioned researchers. However, little is pre-
sented in the reliability study of computer network with mul-
tiple clients and server. The present paper intends to study
performance analysis of computer with multiple clients and
servers by applying supplementary variable technique and
Laplace transforms. Expression of system reliability, avail-
ability, mean time to system failure (MTTF), sensitivity for
MTTF and cost analysis has been computed for various val-
ues of failure and repair rates.
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Life Cycle Reliability and Safety Engineering
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2 Notations, assumptions anddescription
ofthesystem
2.1 Notations
t: Time variable on a time scale.
s: Laplace transform variable for all expressions.
𝜆1
: Failure rate of client (subsystem 1).
𝜆2
: Failure rate of server (subsystem 2).
𝜙(x)
: Repair rate of client (subsystem 1).
𝜙(y)
: Repair rate of server (subsystem 2).
𝜇0(x)
/
𝜇0(y)
: Repair rates for complete failed states.
pi(t)
: The probability that the system is in Si state at
instants for
i
= 0–10.
P(s)
: Laplace transformation of state transition probabil-
ity
p(t)
.
Pi (x, t): The probability that a system is in state Si for
i = 1…10, the system is running under repair and elapse
repair time is (x, t) with repair variable x and time variable t.
Ep(t)
: Expected profit during the time interval [0, t).
K1, K2: Revenue and service cost per unit time
respectively.
S𝜙(x)
: Notation function
S𝜙(x)
=𝜙(x)e
−
x
∫
o
𝜙(x)
dx
with repair
distribution function
𝜙(x)
.
S𝜙
(s
)
: Laplace transforms of
S𝜙(x)
, i.e.,
S
𝜙(s)=
∞
∫
0
e−sx𝜙(x)e
−
x
∫
0
𝜙(x)dx
dx
.
𝜇0(x)
= Cθ (
u1(x)
,
u2(x)
): An expression of joint probabil-
ity (failed state Si to good state S0) according to Gum-
bel–Hougaard family copula is given as
iv. If server 2 failed, then client 1, client 2, client 3 and
client 4 can work or send request to server 1 as shown
in the diagram below.
Figure1a below depict the reliability block diagram of
the proposed client-server system while Fig.1b is the pro-
posed system modelled as series-parallel system.
2.3 State description
Description of the states in Fig.2 below are defined
below:
So
: Perfect state, all subsystems are okay.
S1
/
S8
: Client’s PC failed and is assigned for repair the
system is operative.
S3
/
S9
: After the failure of first client’s PC, second one
failed and is assigned for repair, the system is operative.
S4
/
S10
: Immediately, third client’s PC failed and is
assign for repair, system is operative.
S5
: Complete failed state due to the failure of whole sub-
system 1.
S2
/
S7
: Partially failed state due to the failure of DDS
I of subsystem 2 and is assigned for repair, the system is
operative.
S6
: Complete failed state, due to the failure of DDS II of
subsystem 2.
2.4 Formulation ofmathematical model
By the probability of considerations and continuity of argu-
ments, the following sets of difference differential equations
are associated with the mathematical model.
(1)
(
𝜕
𝜕t+4𝜆1+2𝜆2
)
po(t)=
∞
∫
o
𝜙(x)p1(x,t)dx +
∞
∫
o
𝜙(y)p2(y,t)dy +
∞
∫
0
𝜇0(x)p5(x,t)dx +
∞
∫
0
𝜇0(y)p6(y,t)dy
,
(2)
(𝜕
𝜕t
+
𝜕
𝜕x
+3𝜆1+2𝜆2+𝜙(x)
)
p1(x,t)=
0,
(3)
(
𝜕
𝜕t+𝜕
𝜕y+4𝜆1+𝜆2+𝜙(y)
)
p2(y,t)=
0,
(4)
(𝜕
𝜕t
+
𝜕
𝜕x
+2𝜆1+𝜙(x)
)
p3(x,t)=
0,
(5)
(𝜕
𝜕t
+
𝜕
𝜕x
+𝜆1+𝜙(x)
)
p4(x,t)=
0,
(6)
(𝜕
𝜕t
+
𝜕
𝜕x
+𝜆1+𝜇0(x)
)
p5(x,t)=
0,
c
𝜃
u1(x),u2(x)
=exp
x𝜃+
log 𝜙(x)𝜃
1
𝜃
,
where
𝜃
is a
parameter,
𝜇1=𝜙(x)
, and
u2=ex
and
1≤𝜃≤∞
. For
𝜃=1
,
the Gumbel–Hougaard models’ independence; for
𝜃→∞
,
it converges to comonotonicity.
DDS I: Distributed database server I.
DDS II: Distributed database server II.
2.2 Assumptions
i. It is assumed that server 1 is heavenly loaded server.
ii. It is also assumed that server 2 is lightly loaded server.
iii. If server 1 failed, then client 1 and client 2 cannot
work only client 3 and client 4 can work or send
request to server 2 as shown in the diagram below.
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Life Cycle Reliability and Safety Engineering
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(7)
(
𝜕
𝜕t+𝜕
𝜕y+𝜇0(y)
)
p6(y,t)=
0,
(8)
(
𝜕
𝜕t+𝜕
𝜕y+𝜆2+𝜙(y)
)
p7(y,t)=
0,
(9)
(𝜕
𝜕t
+
𝜕
𝜕x
+3𝜆1+𝜙(x)
)
p8(x,t)=
0,
(10)
(𝜕
𝜕t
+
𝜕
𝜕x
+2𝜆1+𝜙(x)
)
p9(x,t)=
0,
(11)
(𝜕
𝜕t
+
𝜕
𝜕x
+𝜆1+𝜙(x)
)
p10(x,t)=
0.
Boundary conditions
(12)
p1(0, t)=4𝜆1p0(t),
(13)
p2(0, t)=2𝜆2p0(t),
(14)
p3(0, t)=3
𝜆
1p1(0, t),
(15)
p4(0, t)=2𝜆1p3(0, t),
(16)
p5
(0, t)=𝜆
1(
p
4
(0, t)+p
10
(0, t)
),
(17)
p6
(0, t)=𝜆
2(
p
2
(0, t)+p
7
(0, t)
),
2
3
4
1
1
2
Subsystem 1
Subsystem 2
(a)
(b)
Fig. 1 a Reliability block diagram of the system. b Reliability block diagram of the network as series–parallel system
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Life Cycle Reliability and Safety Engineering
1 3
(18)
p7(0, t)=2𝜆2p1(0, t),
(19)
p8(0, t)=4
𝜆
1p2(0, t),
Initial condition
(20)
p9(0, t)=3𝜆1p8(0, t),
(21)
p10(0, t)=2
𝜆
1p9(0, t).
Fig. 2 Reliability transition diagram of mathematical model
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Life Cycle Reliability and Safety Engineering
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2.5 Solution ofthemodel
Taking Laplace transformation of the Eqs.(1)–(21) with help
of Eq.(22) one obtain
(22)
p0(0)=1 and other state transition probabilities are zero at t=0.
(23)
(
s+4𝜆1+2𝜆2
)
p(s)=1+
∞
∫
0
𝜙(x)p1(x,s)dx +
∞
∫
0
𝜙(y)p2(y,s)dy +
∞
∫
0
𝜇0(x)p5(x,s)dx +
∞
∫
0
𝜇0(y)p6(y,s)dy
,
(24)
(
s+
𝜕
𝜕x
+3𝜆1+2𝜆2+𝜙(x)
)
p1(x,s)=
0,
(25)
(
s+𝜕
𝜕y
+4𝜆1+𝜆2+𝜙(y)
)
p2(y,s)=
0,
(26)
(
s+
𝜕
𝜕x
+2𝜆1+𝜙(x)
)
p3(x,s)=
0,
(27)
(
s+
𝜕
𝜕x
+𝜆1+𝜙(x)
)
p4(x,s)=
0,
(28)
(
s+
𝜕
𝜕x
+𝜆1+𝜇0(x)
)
p5(x,y)=
0,
(29)
(
s+𝜕
𝜕y
+𝜇0(y)
)
p6(y,s)=
0,
(30)
(
s+𝜕
𝜕
y
+𝜆2+𝜙(y)
)
p7(y,s)=
0,
(31)
(
s+
𝜕
𝜕x
+3𝜆1+𝜙(x)
)
p8(x,s)=
0,
(32)
(
s+
𝜕
𝜕x
+2𝜆1+𝜙(x)
)
p9(x,s)=
0,
(33)
(
s+
𝜕
𝜕x
+𝜆1+𝜙(x)
)
p10(x,s)=
0,
(34)
p1(0, s)=4𝜆1p0(s),
(35)
p2(0, s)=2𝜆2p0(s),
(36)
p3(0, s)=3
𝜆
1p1(0, s),
Solving Eqs.(23)–(33) with the help of Eqs.(34)–(43),
one may get,
(37)
p4(0, s)=2
𝜆
1p3(0, s),
(38)
p5
(0, s)=𝜆
1(
p
4
(0, s)+p
10
(0, s)
),
(39)
p6
(0, s)=𝜆
2(
p
2
(0, s)+p
7
(0, s)
),
(40)
p7(0, s)=2
𝜆
2p1(0, s),
(41)
p8(0, s)=4𝜆1p2(0, s),
(42)
p9(0, s)=3𝜆1p8(0, s),
(43)
p10(0, s)=2
𝜆
1p9(0, s).
(44)
p
0(s)=
1
D(s),
(45)
p
1(s)=4𝜆1
D(s)
[
1−s𝜙
(
s+3𝜆1+2𝜆2
)
s+3𝜆1+2𝜆2
],
(46)
p
2(s)=2𝜆2
D(s)
[
1−s𝜙
(
s+4𝜆1+𝜆2
)
s+4𝜆1+𝜆2
],
(47)
p
3(s)=12𝜆1
D(s)
[
1−s𝜙
(
s+2𝜆1
)
s+2𝜆1
],
(48)
p
4(s)=
24𝜆3
1
D(s)
[
1−s𝜙
(
s+𝜆1
)
s+𝜆1
],
(49)
p
5(s)=
24𝜆4
1+48𝜆3
1𝜆2
D(s)
1−s𝜇0
(s)
s
,
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Life Cycle Reliability and Safety Engineering
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(50)
p
6(s)=
2𝜆2
2+8𝜆1𝜆2
2
D(s)
1−s𝜇0
(s)
s
,
(51)
p
7(s)=8𝜆1𝜆2
D(s)
[
1−s𝜙
(
s+𝜆2
)
s+𝜆2
],
(52)
p
8(s)=8𝜆1𝜆2
D(s)
[
1−s𝜙
(
s+3𝜆1
)
s+3𝜆1
],
(53)
p
9(s)=24𝜆1𝜆2
D(s)
[
1−s𝜙
(
s+2𝜆1
)
s+2𝜆1
],
where
The Laplace transformations of the state transition prob-
abilities that the system is operative. I.e. perfect and par-
tially failed state (S0, S1, S2, S5, S8, S9, S10) at any time are
as follows:
(54)
p
10(s)=
48𝜆2
1𝜆2
D(s)
[
1−s𝜙
(
s+𝜆1
)
s+𝜆1
],
(55)
D
(s)=s+4𝜆1+2𝜆2−
4𝜆1
s
𝜙
s+3𝜆1+2𝜆2
+
2𝜆2s
𝜙s+4𝜆1+𝜆2+
24𝜆4
1+48𝜆3
1𝜆2
+
2𝜆2
2
+8𝜆1𝜆2
2
+
s𝜇0(s)
.
(56)
p
up
(
s
)=
p0
(
s
)+
p1
(
s
)+
p2
(
s
)+
p3
(
s
)+
p4
(
s
)
+
p7
(
s
)+
p8
(
s
)+
p9
(
s
)+
p10
(
s
)
,
(57)
=
1
D(s)
1+4𝜆1
1−s𝜙
s+3𝜆1+2𝜆2
s+3𝜆1+2𝜆2
+2𝜆2
1−s𝜙
s+4𝜆1+𝜆2
s+4𝜆1+𝜆2
+
12𝜆2
11−s𝜙s+2𝜆1
s+2𝜆1+24𝜆3
11−s𝜙1s+𝜆1
s+𝜆1+
8𝜆1𝜆21−s𝜙s+𝜆2
s+𝜆2+8𝜆1𝜆21−s𝜙s+3𝜆1
s+3𝜆1+
24𝜆1𝜆21−s𝜙s+2𝜆1
s+2𝜆1+48𝜆2
1𝜆21−s𝜙s+𝜆1
s+𝜆1
,
(58)
pdown(s)=1−pup (s).
Table 1 Variation of availability with respect of time when the repair follows copula distribution
Time 0 4 8 12 16 20 24 28 32 36 40
Availability 1.000 0.996 0.982 0.967 0.952 0.938 0.924 0.910 0.897 0.883 0.870
Table 2 Variation of availability with respect of time when the repair follows general distribution
Time 0 4 8 12 16 20 24 28 32 36 40
Availability 1.000 0.994 0.976 0.958 0.941 0,924 0.907 0.891 0.875 0.859 0.843
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Life Cycle Reliability and Safety Engineering
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3 Analytical study ofthemodel asparticular
cases
3.1 Availability analysis withcopula distribution
If the repair follows Gumbel–Hougaard family copula dis-
tribution then, setting.
S
𝜇0(s)=̄
Sexp[x𝜃+{log 𝜑(x)}𝜃]1∕𝜃(s)= exp[x
𝜃
+{log 𝜑(x)}
𝜃
]
1∕𝜃
s+exp[x
𝜃
+{log 𝜑(x)}
𝜃
]
1∕𝜃 ,
̄
S
𝜙(s)=
𝜙
s+𝜙,
and taking the values of different parameters
as λ1 = 0.01, λ2 = 0.02,
𝜙=𝜇=x=y=1
and
𝜙(x)=𝜙(y)=1
in Eq.(57), then taking inverse Laplace
transform, one can obtain the availability expression as:
(59)
p
up(s)=
⎡
⎢
⎢
⎣
−0.002772e−1.02000t+0.000319e−2.71917t−0.009263e−1.14065t
−0.000063e−1.06469t+1.012316e−0.00377t−0.000050e−.01000t
−0.000487e−1.03000t
⎤
⎥
⎥
⎦
.
For different values of time variable t = 0, 4, 8, 12…40,
units of time, one may get different values of Availability
with the help of Eq.(59) as shown in Table1.
3.2 Availability analysis withgeneral distribution
If the repair follows general distribution then, setting,
S
𝜙(s)=
𝜙
s+𝜙
, and taking the values of different values of fail-
ure and repair rates parameters as λ1 = 0.01, λ2 = 0.02,
𝜙=𝜇=x=y=1
in Eq. (57), and then taking inverse
Laplace transform, one can obtain the expression for avail-
ability as:
For different values of time variable t = 0, 4, 8, 12…40,
units of time, one may get different values of availability
with the help of Eq.(60) as shown in Table2.
(60)
p
up(s)=
[
−0.000487e
−
1.03000t−0.002773e
−
1.02000t−0.000050e
−
1.01000t
−0.009192e−1.14074t−0.0000626e−1.06469t+1.0125660e−0.00455t
].
Table 3 Computation of
reliability for different values of
time (t)
Time 0 4 8 12 16 20 24 28 32 36 40
Reliability 1.000 0.996 0.936 0.847 0.749 0.650 0.558 0.474 0.400 0.336 0.282
Table 4 Computation of MTTF in respect to failure rates
Failure rate MTTF
𝜆1
MTTF
𝜆2
0.01 31.709 41.564
0.02 23.351 31.709
0.03 19.494 26.050
0.04 17.228 22.265
0.05 15.732 19.513
0.06 14.674 17.403
0.07 13.890 15.726
0.08 13.289 14.356
0.09 12.818 13.215
Table 5 Sensitivity MTTF in respect of failure rates
Failure rate
𝜕(MTTF)
𝜆
1
𝜕(MTTF)
𝜆
2
0.01 − 1356.535 − 1377.481
0.02 − 523.481 − 713.041
0.03 − 285.055 − 451.710
0.04 − 180.276 − 318.027
0.05 − 124.069 − 238.599
0.06 − 90.181 − 186.763
0.07 − 68.097 − 150.714
0.08 − 52.874 − 124.468
0.09 − 41.924 − 104.685
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Life Cycle Reliability and Safety Engineering
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3.3 Reliability analysis
Taking all repair rates,
𝜙
,
𝜇0
, in Eq.(57) to zero for some values
of failure rates as λ1 = 0.01, λ2 = 0.02 and then taking Laplace
transformation, one can get reliability expression as:
The computation of R(t) for t = 0, 4, 8, 12, 16, 20, 24, 28,
32, 36, 40 is depicted in Table3.
3.4 Mean time tofailure (MTTF) analysis
The mean time between failures (MTBF) is the predicted
elapsed time between inherent system, during normal system
operation. The MTBF can be calculated as the arithmetic
mean (average) time between failures of a system. If R(t) is
the reliability function obtained by taking inverse Laplace
(61)
R
(t)=
[
−5.160380e
−0.08000t
+4e
−0.07000t
+2e
−0.06000t
+0.126666e
−0.02000t
+0.001714e−0.01000t+0.032000e−0.03000t
].
transform of Pup(s) than average time to system failure for a
continuous-valued function,
MTTF =
∫∞
0R(t)dt =lim
s→0
R(s)
.
Taking all repairs to zero in equation in (57) and the limit
as s tends to zero one can obtain the expression for MTTF
as:
Setting λ1 = 0.01, λ2 = 0.02 and changing λ1, λ2, one by
one respectively as 0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07,
0.08, 0.09 in Eq.(62), one may obtain the variation of
M.T.T.F. with respect to failure rates as shown in Table4.
(62)
MTTF
=lim
s→0pup(s)=1
4𝜆1+2𝜆2
⎡
⎢
⎢
⎢
⎣
1+
4𝜆
1
3𝜆1+2𝜆2
+
2𝜆
2
4𝜆1+𝜆2
+
14𝜆1+24𝜆2
1+44𝜆2
3
+48𝜆1𝜆2
⎤
⎥
⎥
⎥
⎦
.
Table 6 Expected profit in [0, t)
t = 0, 4, 8…40, when the repair
follows copula distribution
Time (t)
Ep(t)
K = 0.6
Ep(t)
K = 0.5
Ep(t)
K = 0.4
Ep(t)
K = 0.3
Ep(t)
K = 0.2
Ep(t)
K = 0.1
0 000000
4 1.609 2.007 2.407 2.807 3.207 3.607
8 3.165 3.965 4.765 5.565 6.365 7.165
12 4.664 5.864 7.064 8.264 9.464 10.664
16 6.105 7.705 9.305 10.905 12.505 14.105
20 7.488 9.488 11.488 13.488 15.488 17.488
24 8.814 11.214 13.614 16.014 18.414 20.814
28 10.085 12.885 15.685 18.485 21.285 24.085
32 11.300 14.500 17.700 20.900 24.100 27.300
36 12.461 16.061 19.661 23.261 26.861 30.461
40 13.569 17.569 21.569 25.569 29.569 33.569
Table 7 Expected profit in [0, t)
t = 0, 4, 8…40. When the repair
follows general distribution
Time (t)
Ep(t)
K = 0.6
Ep(t)
K = 0.5
Ep(t)
K = 0.4
Ep(t)
K = 0.3
Ep(t)
K = 0.2
Ep(t)
K = 0.1
0 000000
4 1.601 2.000 2.401 2.801 3.201 3.601
8 3.142 3.942 4.742 5.542 6.342 7.142
12 4.612 5.812 7.012 8.212 9.412 10.612
16 6.012 7.612 9.212 10.812 12.412 14.012
20 7.343 9.343 11.343 13.343 15.343 17.343
24 8.607 11.007 13.407 15.807 18.207 20.607
28 9.805 12.605 15.405 18.205 21.005 23.805
32 10.938 14.138 17.338 20.538 23.738 26.938
36 12.006 15.606 19.206 22.806 26.406 30.006
40 13.013 17.013 21.013 25.013 29.013 33.013
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3.5 Sensitivity analysis
The sensitivity in MTTF of the system computed through
the partial differentiation of MTTF with respect to the fail-
ure rates of the system. By applying the set of parameters
as, λ1 = 0.01, λ2 = 0.02, and in the partial differentiation of
MTTF, one can calculate the MTTF sensitivity as shown
in Table5 below.
3.6 Cost analysis withcopula distribution
If the service facility is always available, then expected
profit during the interval [0, t] is give by the formula with
service facility K1 and K2 service cost per unit time in the
interval [0, t).
For the same set of the parameters of Eq.(59), one can
obtain Eq.(64) as:
(63)
E
p(t)=K1
t
∫
0
Pup(t)dt −K2t
.
(64)
E
p(t)=k1
⎡
⎢
⎢
⎢
⎣
0.002718e
−0.02000t
−0.000173e
−2.21917t
+0.008120e−1.14065t+0.000059e−1.06469t
−267.872690e0.00377t+0.000049e−1.01000t
+0.0004729e
−1.03000t
+267.861
⎤
⎥
⎥
⎥
⎦
−k2(t)
.
Fig. 3 Availability against time
when the repair follows copula
distribution
0.86
0.88
0.9
0.92
0.94
0.96
0.98
1
1.02
0 10 20 30 40 50
Availability
Time (t)
Availability
Fig. 4 Availability against time
when the repair follows general
distribution
0.82
0.84
0.86
0.88
0.9
0.92
0.94
0.96
0.98
1
1.02
0 5 10 15 20 25 30 35 40 45
Availability
Time (t)
Availability
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Life Cycle Reliability and Safety Engineering
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Setting K1 = 1 and K2 = 0.6, 0.5, 0.4, 0.3, 0.2 and 0.1
respectively and changing t = 0, 4, 8, 12, 16, 20, 24, 28,
32, 36, 40. Units of time, the results for expected profit can
be obtain as shown in Table6.
3.7 Cost analysis withgeneral distribution
If the service facility is always available, then expected
profit during the interval [0, t] is give by the formula with
service facility K1 and K2 service cost per unit time in the
interval [0, t).
For the same set of the parameters of Eq.(59), one can
obtain Eq.(66) as:
(65)
E
p(t)=K1
t
∫
0
Pup(t)dt −K2t
.
(66)
E
p(t)=k1
⎡
⎢
⎢
⎢
⎣
0.000473e
−0.03000t
+0.002719e
−1.02000t
+0.000049e−1.01000t+0.008058e−1.14074t
+0.000058e−1..06469t−222.147834e−.00455t
+222.136
⎤
⎥
⎥
⎥
⎦
−k2(t)
.
Fig. 5 Reliability against time
0
0.2
0.4
0.6
0.8
1
1.2
0 10 20 30 40 50
Reliability
Time (t)
Reliability
Fig. 6 MTTF against failure
rate
0
5
10
15
20
25
30
35
40
45
0 0.02 0.04 0.06 0.08 0.1
MTTF
Failure Rate
MTTF
MTTF
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Life Cycle Reliability and Safety Engineering
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Setting K1 = 1 and K2 = 0.6, 0.5, 0.4, 0.3, 0.2 and 0.1
respectively and varying t = 0, 4, 8, 12, 16, 20, 24, 28, 32,
34, 40. Units of time, the results for expected profit can be
obtain as shown in Table7.
4 Discussion
Table1 and Fig.3 depict the availability of the system with
respect to the time when repairs follow copula distribution
when other parameters are fixed. It is evident from the table
and simulation in Fig.3 that availability decrease smoothly
with passage of time. On the other hand when repairs follow
general distribution, Table2 and simulation Fig.4 displayed
the effect of passage time on system availability when other
parameters are kept constant. It is clear from the Table2 and
Fig.4 that availability decreases smoothly with passage of
time. However from, availability tends to be higher when
repairs follows copula distribution than general distribution.
Table3 and Fig.5 displayed the trend of reliability with
passage of time when other parameters are kept constant. It
is clear from the table and figure that reliability decreases
with passage of time. Reliability tends to decrease when
the passage time is around t = 4 and thereafter it decreases
smoothly as passage time increases.
Fig. 7 Sensitivity against failure
rate
-1600
-1400
-1200
-1000
-800
-600
-400
-200
0
0 2 4 6 8 10 12
Sensivity
Failure Rate
Series2
Series3
Fig. 8 Expected profit against
time, when the repair follows
copula distribution
0
5
10
15
20
25
30
35
40
0 10 20 30 40 50
Expected Profit
Time (t)
K = 0.6
K = 0.5
K = 0.4
K = 0.3
K = 0.2
K = 0.1
Author's personal copy
Life Cycle Reliability and Safety Engineering
1 3
Table4 and Fig.6 displayed the trend of mean time to
failure with the failure rates
𝜆1
and
𝜆2
when other param-
eters are kept constant. It is clear from the table and fig-
ure that the mean time to failure decreases as
𝜆1
and
𝜆2
increases. Variation of mean time to failure become closer
as
𝜆1
and
𝜆2
increases as depicted in the table.
Table5 and Fig.7 displayed the variation of sensitivity
analysis with respect to the failure rates
𝜆1
and
𝜆2
when
other parameters are kept constant. It is clear from the
table and figure that the sensitivity increases as
𝜆1
and
𝜆2
increases. Variation of sensitivity become closer as
𝜆1
and
𝜆2
increases as depicted in the table.
Table6 and Fig.8 displayed the trend of profit with
respect to the time for different values of service cost when
repairs follow copula distribution when other parameters
are fixed. It is evident from Table6 and simulation in
Fig.8 that the profit increase with passage of time for dif-
ferent service cost. However, the profit in higher for when
service K = 0.1 as passage time increases. Similarly when
repairs follows general distribution in Table7 and simula-
tion Fig.9, the effect of passage time on profit is closer
when to the case when repairs follows copula distribution
other parameters are kept constant. However, the profit
tends to be higher when repairs follow copula distribution
than general distribution as shown in Figs.9.
5 Concluding remark
In the present paper, a computer network system is con-
sidered in which the network consist of four homogeneous
clients joined by two homogeneous servers. Both the cli-
ents and servers are configured as series parallel systems
as depicted in Fig.1b in the paper. Expressions for system
reliability, availability, mean time to failure as well as cost
function are derived. Impact of passage time, failure rate
of reliability measures have been presented in the form
tables and graphs. It is evident from the analysis presented
that network reliability and availability can be enhance
by maintaining failure free scenario, invoking preventive
maintenance measure in order to keep the system away
from unnecessary failure. The study has shown multiple
server can be employed in enhancing the system reliability.
In future research, distributed hardware software network
will be studied in which each consist of two heterogeneous
software saving similar purpose.
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